# Solve the given simultaneous equation

• chwala
Reals between 0 and 1In summary, the conversation is about solving for x and y in a simultaneous equation problem. The initial approach used an algebraic method, while the other conversation participants explored alternative approaches such as the quartic equation method and numerical methods. Ultimately, the quartic equation method was used and the approximate value of x was found to be 2. The conversation also touched on the concept of complex roots in polynomials and the estimation that at least half of polynomials have complex roots.
chwala
Gold Member
Homework Statement
Solve for ##x## and ##y## given;
##\sqrt x+ y=7##
##x+\sqrt y=11##
Relevant Equations
Simultaneous equations
I saw this problem on you tube ... The working to solution is much clear to me. Find the link here;

I am now trying to look at an alternative approach. My thinking is as follows;
##\sqrt x=7-y##
##x=(7-y)^2##

##(7-y)^2+\sqrt y=11##
##49-14y+y^2+\sqrt y=11##
i then let ##\sqrt y=m## giving me,
##m^4-14m^2+m+38=0## at this point i am now exploring ways of solving quartic equations...looking at it...then will respond on my finding...i hope i am on the right path...

Delta2
What is the point of this thread?

drmalawi said:
What is the point of this thread?
Am I on the right path? My post is clear. I am exploring on alternative approach.

chwala said:
My post is clear
There was no question.

I can give you a hint regarding the quartic equation. Can you find a simple positive m-value which will solve it?
Also, you can with derivative method figure out how many positive roots that quartic equation will have.
You can also use that ##m < 7## to pinpoint down the only solution you care about in that quartic equation.

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Delta2 and chwala
Yes, I am looking into this ...looking at the numerical methods...I'll get back once done...

I just solved it following your work with no aid of any calculator.
Was kinda fun :)

chwala
chwala said:
I am now trying to look at an alternative approach.
EDITED

Is an algebraic solution essential? Can I point out (if not already evident) that the problem can be solved using some 'guesswork'.

Click on the spoiler below for hints...

From inspection:
- x and y look likely to be not only integers but also perfect squares; we can try this to see where it leads
- values of 0 or 1 for x and y are easily excluded;
- y must be less than 6 giving a solution for y.

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chwala
I noticed a mistake in the video...on the factorisation part; ought to be
...
##(\sqrt x-\sqrt y)(1-\sqrt x+\sqrt y)=-4## ...

And it looks like trial and error approach as the method on the video assumes solutions then works backwards to determine the value of variables ##x## and ##y##. I would rather a step-step working to solution. I will look at this today...

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There are various techniques in solving quartic equations; phew a bit technical...Ferrari method, Durand- Kerner method ...need to read on them.

Delta2
You don't need those

This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...

chwala
Delta2 said:
This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...
I have seen all the ##4## roots with help of calculator...
ie ##-3.2831859, -1.8481265, 2 ## and ## 3.1313125##

Delta2
drmalawi said:
I just solved it following your work with no aid of any calculator.
Was kinda fun :)
Arrrrrrgh! I can now see that we can get the roots by using Factor theorem! Eeeeeeeeish!

Delta2
chwala said:
Arrrrrrgh! I can now see that we can get the roots by using Factor theorem! Eeeeeeeeish!
I wrote that in post #4, don't you read the replies?

You first try with some solutions, you must have that 0 < m2 < 7, so try with some solutions for m.
You will find pretty quick that m = 2 works.
Then do polynomial division to get a cubic polynomial.
Show that the resulting polynomial does not equal zero for any positive m less than sqrt(7), I used some basic curve sketch analysis with derivative method.

chwala
Delta2 said:
This polynomial we end up with $$m^4-14m^2+m+38$$ is very interesting, it has 4 roots ALL real...
What is so interesting with a quartic polynomial with four real roots?

Delta2
drmalawi said:
What is so interesting with a quartic polynomial with four real roots?
Well I don't know to be honest, for some reason my intuition told me that most polynomials of high degree have at least one complex root.

I got it! I will post my step-step solution later in the day...I used the quartic approach- converted to cubic terms...with some algorithm involved. Bingo talk later...

drmalawi said:
I wrote that in post #4, don't you read the replies?

You first try with some solutions, you must have that 0 < m2 < 7, so try with some solutions for m.
You will find pretty quick that m = 2 works.
Then do polynomial division to get a cubic polynomial.
Show that the resulting polynomial does not equal zero for any positive m less than sqrt(7), I used some basic curve sketch analysis with derivative method.
This approach is straightforward man...i will post the approach that is used generally for quartic equations ..cheers

Delta2 said:
most polynomials of high degree have at least one complex root.
How do you compare the amounts of polynomals to quantify "most"?

chwala said:
This approach is straightforward man.
Yes it is, I give my 16-17y old students similar problems which are solved using this approach.

chwala
drmalawi said:
How do you compare the amounts of polynomals to quantify "most"?
Pure intuition, there are infinite polynomials but my intuition told me at at least half of them have a complex root.

chwala
chwala said:
Homework Statement:: Solve for ##x## and ##y## given;
##\sqrt x+ y=7##
##x+\sqrt y=11##
Relevant Equations:: Simultaneous equations

I saw this problem on you tube ... The working to solution is much clear to me. Find the link here;

I am now trying to look at an alternative approach. My thinking is as follows;
##\sqrt x=7-y##
##x=(7-y)^2##

##(7-y)^2+\sqrt y=11##
##49-14y+y^2+\sqrt y=11##
i then let ##\sqrt y=m## giving me,
##m^4-14m^2+m+38=0## at this point i am now exploring ways of solving quartic equations...looking at it...then will respond on my finding...i hope i am on the right path...

We have;
##m^4-14m^2+m+38=0##

On Using the approach given on this link https://www.1728.org/quartic2.htm

We shall have;
##f=-14##
##g=1##
##h=38##

Our cubic equation is;
##y^3-7y^2+\dfrac{44}{16}y-\dfrac{1}{64}=0##

On solving, we have

##y_1=6.583##
##y_2=0.412##
##y_3=0.0058##

giving us;

##p=2.566##
##q=0.6419##
##r=-0.0758##

substituting into the equations, we end up with;

##x_1=p+q+r-s=3.132##
##x_2=-p+q-r-s=-1.8483##
##x_3=-p-q+r-s=-3.2837##
##x_4=p-q-r-s=1.999999≅2## which is the envisaged value that we are looking for...from here the working to solution will follow...

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Delta2
chwala said:
##x_4=p-q-r-s=1.999999≅2##
how come you don't get exactly 2?

drmalawi said:
how come you don't get exactly 2?
The approach is an approximate method (Numerical method) and not exact.

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Delta2
Delta2 said:
Pure intuition, there are infinite polynomials but my intuition told me at at least half of them have a complex root.
There are uncountably-many polynomials over the Reals, uncountably many with Complex roots( degree 2 or higher) ; those that are multiples of ##x^2+a ; a>0## alone are enough*, and uncountably-many with Real roots: ## ( x-r_1)(x-r_2)(x-r_3)(x-r_4); r_i \in \mathbb R##
*Similar argument if you want all roots to be Real.

chwala and Delta2
I'd suggest in general to get a feel for it, as someone had suggested. And, maybe too, a more geometric approach using parabolas ( after squaring; set ## \sqrt u:= u, etc), though you'd have to go back-forth in the change of variables.

chwala
WWGD said:
There are uncountably-many polynomials over the Reals, uncountably many with Complex roots( degree 2 or higher) ; those that are multiples of ##x^2+a ; a>0## alone are enough*, and uncountably-many with Real roots: ## ( x-r_1)(x-r_2)(x-r_3)(x-r_4); r_i \in \mathbb R##
*Similar argument if you want all roots to be Real.
Hmm, can we say that the probability a random polynomial to have only real roots is 50%? Hard though to imagine how we define this probability as if we try to define it as a fraction, both the numerator and the denominator are uncountable many...

I made wolfram draw graphs of y=11-x^2 and x=7-y^2 where ##\sqrt{x},\sqrt{y}## in the problem statement is transferred to x, y.

I could confirm that there is only one solution for x,y>0, obvious x=3, y=2.

Equation for x is
$$(11-x^2)^2=7-x$$
$$x^4-22x^2+x+114=0$$
$$(x-3)(x^3+3x^2-13x-38)=0$$

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chwala and Delta2
malawi_glenn said:
How do you compare the amounts of polynomals to quantify "most"?Yes it is, I give my 16-17y old students similar problems which are solved using this approach.
Descarte's Rule Of Signs.

MidgetDwarf said:
Descarte's Rule Of Signs.
Yeah but we do this in context with derivative of polynomial functions.
But thanks for the reminder! That is very useful trick.

Delta2 said:
Hmm, can we say that the probability a random polynomial to have only real roots is 50%?
Not without specifying a probability distribution on the polynomials. The result will depend on the distribution you impose.

Since both sets are aleph one, it is possible to construct a bijection between them.

chwala and malawi_glenn
Orodruin said:
Since both sets are aleph one, it is possible to construct a bijection between them.
I was thinking about that too. I recalled a problem in an real analysis book of mine which I did not read that long time ago which asked how many algebraic vs. trancendetal numbers (over the rationals) there are. If I remember correctly, there should be countable many (##\aleph_0##) algebraic numbers and uncountable (##\aleph_1##) trancendental numbers... I should revisit my notes at some points

Delta2 said:
Hmm, can we say that the probability a random polynomial to have only real roots is 50%? Hard though to imagine how we define this probability as if we try to define it as a fraction, both the numerator and the denominator are uncountable many...
I guess we'd have to choose a distribution to determine probabilities.

chwala and Delta2
chwala said:
Homework Statement:: Solve for ##x## and ##y## given;
##\sqrt x+ y=7##
##x+\sqrt y=11##
Relevant Equations:: Simultaneous equations
Presuming this is a Diophantine (integer only) equation , we don't have too many choices:

1 + 6
2 + 5
3 + 4

One number has to be a perfect square. Exclude 2 + 5. If x=1, y=6 and that doesn't work either.

chwala and malawi_glenn

## 1. What is a simultaneous equation?

A simultaneous equation is a set of two or more equations that are solved together to find the values of the variables that satisfy all of the equations.

## 2. How do I solve a simultaneous equation?

To solve a simultaneous equation, you can use various methods such as substitution, elimination, or graphing. These methods involve manipulating the equations to eliminate one variable and solve for the other.

## 3. Can a simultaneous equation have more than two variables?

Yes, a simultaneous equation can have any number of variables. However, the number of equations should be equal to the number of variables in order to find a unique solution.

## 4. What is the importance of solving simultaneous equations?

Solving simultaneous equations is important in many fields of science and mathematics. It allows us to find the values of multiple variables that satisfy a set of equations, which can be used to make predictions, solve real-world problems, and model complex systems.

## 5. Can all simultaneous equations be solved?

No, not all simultaneous equations have a unique solution. Some equations may have no solution, while others may have an infinite number of solutions. It depends on the specific equations and variables involved.

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